💡Okay, so I have this inequality: (3x^2 - log_a x < 0), and it needs to hold true for any (x) in the interval ((0, frac{1}{3})). I need to find the range of the real number (a). The options are given as A, B, C, D, with different intervals. Hmm, let me try to figure this out step by step.First, I know that logarithms can be tricky because their behavior depends on the base (a). If (a > 1), the logarithm function (log_a x) is increasing, and if (0 < a < 1), it's decreasing. Also, when (x) is between 0 and 1, (log_a x) will be positive if (a < 1) and negative if (a > 1). Wait, is that right? Let me think. If (a > 1), then (log_a x) is negative because (x < 1). If (0 < a < 1), then (log_a x) is positive because (x < 1). Yeah, that seems correct.So, the inequality is (3x^2 - log_a x < 0). Let me rewrite that as (3x^2 < log_a x). So, depending on the value of (a), (log_a x) can be positive or negative. Let me consider two cases: when (a > 1) and when (0 < a < 1).**Case 1: (a > 1)**In this case, as I thought earlier, (log_a x) is negative because (x) is between 0 and (frac{1}{3}), which is less than 1. So, the right side of the inequality (3x^2 < log_a x) is negative. But the left side, (3x^2), is always positive because (x^2) is positive and multiplied by 3. So, we have a positive number less than a negative number, which is impossible. Therefore, the inequality cannot hold for (a > 1). So, (a) cannot be greater than 1. That rules out any options where (a) is greater than 1, but looking at the options, all of them have (a) less than or equal to 1, so maybe I need to consider the other case.**Case 2: (0 < a < 1)**Here, (log_a x) is positive because (x < 1) and (a < 1). So, both sides of the inequality (3x^2 < log_a x) are positive. That makes more sense because we can have a positive number less than another positive number.I need this inequality to hold for all (x) in ((0, frac{1}{3})). So, I need to find the values of (a) such that (3x^2 < log_a x) for all (x) in that interval.Let me try to manipulate the inequality to solve for (a). Starting with:(3x^2 < log_a x)I can rewrite the logarithm in terms of natural logarithm to make it easier:(log_a x = frac{ln x}{ln a})So, the inequality becomes:(3x^2 < frac{ln x}{ln a})Since (0 < a < 1), (ln a) is negative. So, when I multiply both sides by (ln a), the inequality sign will reverse. Let's do that:(3x^2 cdot ln a > ln x)So,(3x^2 cdot ln a > ln x)Hmm, this seems a bit complicated. Maybe I can rearrange it differently. Let me think about the function (f(x) = 3x^2 - log_a x). We need (f(x) < 0) for all (x in (0, frac{1}{3})). So, the maximum of (f(x)) in that interval should be less than 0.To find the maximum, I can take the derivative of (f(x)) and set it equal to zero.(f(x) = 3x^2 - log_a x)First, find the derivative (f'(x)):(f'(x) = 6x - frac{1}{x ln a})Set (f'(x) = 0):(6x - frac{1}{x ln a} = 0)Multiply both sides by (x ln a):(6x^2 ln a - 1 = 0)So,(6x^2 ln a = 1)(ln a = frac{1}{6x^2})Therefore,(a = e^{frac{1}{6x^2}})Hmm, interesting. So, the critical point occurs at (a = e^{frac{1}{6x^2}}). But I'm not sure if this is helpful yet. Maybe I should instead consider the behavior of (f(x)) at the endpoints of the interval.Wait, but (x) approaches 0 from the right, so let's see what happens as (x) approaches 0.As (x to 0^+):- (3x^2 to 0)- (log_a x to infty) because (x to 0) and (a < 1), so (log_a x) tends to infinity.Therefore, (f(x) = 3x^2 - log_a x to -infty), which is less than 0. So, near 0, the inequality holds.At (x = frac{1}{3}):(fleft(frac{1}{3}right) = 3 left(frac{1}{3}right)^2 - log_a left(frac{1}{3}right))Simplify:(3 times frac{1}{9} - log_a left(frac{1}{3}right) = frac{1}{3} - log_a left(frac{1}{3}right))We need this to be less than 0:(frac{1}{3} - log_a left(frac{1}{3}right) < 0)So,(log_a left(frac{1}{3}right) > frac{1}{3})But (log_a left(frac{1}{3}right) = frac{ln left(frac{1}{3}right)}{ln a} = frac{-ln 3}{ln a})Since (0 < a < 1), (ln a < 0), so:(frac{-ln 3}{ln a} > frac{1}{3})Multiply both sides by (ln a), which is negative, so inequality reverses:(-ln 3 < frac{1}{3} ln a)Multiply both sides by 3:(-3 ln 3 < ln a)Exponentiate both sides:(e^{-3 ln 3} < a)Simplify:(e^{ln 3^{-3}} = 3^{-3} = frac{1}{27})So,(frac{1}{27} < a)But we already have (0 < a < 1), so combining these, we get:(frac{1}{27} < a < 1)Wait, but let me check if this is correct. So, at (x = frac{1}{3}), we have the inequality (log_a left(frac{1}{3}right) > frac{1}{3}), which leads to (a > frac{1}{27}). So, (a) must be greater than (frac{1}{27}).But we also need to ensure that the inequality holds for all (x) in ((0, frac{1}{3})), not just at (x = frac{1}{3}). So, maybe I should check if the maximum of (f(x)) occurs at (x = frac{1}{3}) or somewhere else.Earlier, I found that the critical point is at (a = e^{frac{1}{6x^2}}). Let me see if this critical point is within the interval ((0, frac{1}{3})).Wait, actually, the critical point is a function of (x), but (a) is a constant. So, perhaps I need to consider whether the function (f(x)) has a maximum in the interval ((0, frac{1}{3})).Let me analyze the derivative again:(f'(x) = 6x - frac{1}{x ln a})Set (f'(x) = 0):(6x = frac{1}{x ln a})Multiply both sides by (x ln a):(6x^2 ln a = 1)So,(ln a = frac{1}{6x^2})Since (x in (0, frac{1}{3})), (x^2 < frac{1}{9}), so (frac{1}{6x^2} > frac{1}{6 times frac{1}{9}} = frac{9}{6} = frac{3}{2}). Therefore, (ln a > frac{3}{2}), which would imply (a > e^{3/2}), but (e^{3/2}) is approximately 4.48, which contradicts our earlier case where (0 < a < 1). So, this suggests that there is no critical point within the interval ((0, frac{1}{3})) because the equation (6x^2 ln a = 1) cannot be satisfied for (0 < a < 1) and (x in (0, frac{1}{3})). Therefore, the function (f(x)) is either always increasing or always decreasing in the interval.Let me check the sign of the derivative (f'(x)) in the interval. Since (0 < a < 1), (ln a < 0). So, the second term (- frac{1}{x ln a}) becomes positive because (ln a) is negative. So, (f'(x) = 6x + frac{1}{x |ln a|}). Both terms are positive, so (f'(x) > 0) for all (x in (0, frac{1}{3})). That means (f(x)) is increasing on this interval.Since (f(x)) is increasing on ((0, frac{1}{3})), its maximum occurs at (x = frac{1}{3}). Therefore, if (fleft(frac{1}{3}right) < 0), then (f(x) < 0) for all (x in (0, frac{1}{3})).We already found that (fleft(frac{1}{3}right) < 0) implies (a > frac{1}{27}). So, combining this with (0 < a < 1), we get (frac{1}{27} < a < 1).But wait, let me check if (a = frac{1}{27}) is included. If (a = frac{1}{27}), then at (x = frac{1}{3}):(fleft(frac{1}{3}right) = frac{1}{3} - log_{1/27} left(frac{1}{3}right))Calculate (log_{1/27} left(frac{1}{3}right)):(log_{1/27} left(frac{1}{3}right) = frac{ln left(frac{1}{3}right)}{ln left(frac{1}{27}right)} = frac{-ln 3}{-3 ln 3} = frac{1}{3})So,(fleft(frac{1}{3}right) = frac{1}{3} - frac{1}{3} = 0)But the inequality is strict: (f(x) < 0). Therefore, at (a = frac{1}{27}), the inequality is not satisfied at (x = frac{1}{3}) because it equals zero. So, (a) must be strictly greater than (frac{1}{27}).However, looking back at the options, option A is ([frac{1}{27}, 1]), which includes (a = frac{1}{27}), but we just saw that at (a = frac{1}{27}), the inequality fails at (x = frac{1}{3}). Therefore, (a) must be greater than (frac{1}{27}), so the interval is ((frac{1}{27}, 1)), which is option B.Wait, but let me double-check. If (a = frac{1}{27}), then for (x) approaching (frac{1}{3}) from the left, (f(x)) approaches 0 from below because (f(x)) is increasing. So, for (x) just less than (frac{1}{3}), (f(x)) is slightly less than 0. But at (x = frac{1}{3}), it's exactly 0. So, the inequality (f(x) < 0) holds for all (x in (0, frac{1}{3})), but not including (x = frac{1}{3}). Since the interval is open, (x) doesn't include (frac{1}{3}), so maybe (a = frac{1}{27}) is acceptable?Wait, no. Because the inequality has to hold for all (x) in ((0, frac{1}{3})). If (a = frac{1}{27}), then as (x) approaches (frac{1}{3}), (f(x)) approaches 0, but it's still less than 0 for all (x) in ((0, frac{1}{3})). So, maybe (a = frac{1}{27}) is acceptable because the inequality is strict and doesn't include the endpoint.Hmm, this is a bit confusing. Let me think again. The interval for (x) is open, so (x) never actually reaches (frac{1}{3}). Therefore, even if (f(x)) approaches 0 as (x) approaches (frac{1}{3}), it never actually equals 0 in the interval. So, for (a = frac{1}{27}), (f(x) < 0) for all (x in (0, frac{1}{3})). Therefore, (a = frac{1}{27}) is acceptable.Wait, but earlier when I plugged in (a = frac{1}{27}) and (x = frac{1}{3}), I got (f(x) = 0), but since (x = frac{1}{3}) is not included in the interval, it's okay. So, (a = frac{1}{27}) is acceptable because the inequality holds for all (x) in the open interval.Therefore, the range of (a) is (frac{1}{27} leq a < 1). Wait, but looking at the options, A is ([frac{1}{27}, 1]), which includes 1, but earlier we saw that (a) cannot be equal to 1 because when (a = 1), the logarithm is undefined (since (log_1 x) is not defined). So, actually, (a) must be less than 1.Wait, no. If (a = 1), (log_1 x) is undefined because any power of 1 is always 1, so it's not possible to have (log_1 x) for (x neq 1). Therefore, (a) cannot be equal to 1. So, the upper bound is (a < 1). Therefore, the interval is (frac{1}{27} leq a < 1), which is option A: ([frac{1}{27}, 1]), but actually, since (a) cannot be 1, it's ([frac{1}{27}, 1)), but that's not one of the options. Wait, the options are A: ([frac{1}{27}, 1]), B: ((frac{1}{27}, 1)), C: ((0, frac{1}{27})), D: ((0, frac{1}{27}]).Hmm, so since (a) cannot be 1, the correct interval should be ((frac{1}{27}, 1)), which is option B. But earlier, I thought that (a = frac{1}{27}) is acceptable because the inequality holds for all (x) in the open interval. So, is (a = frac{1}{27}) included or not?Wait, let's test (a = frac{1}{27}) with a value just less than (frac{1}{3}), say (x = frac{1}{3} - epsilon) where (epsilon) is a very small positive number. Then, (f(x) = 3x^2 - log_{1/27} x). As (x) approaches (frac{1}{3}), (f(x)) approaches 0 from below, so for (x) just less than (frac{1}{3}), (f(x)) is slightly less than 0. Therefore, (a = frac{1}{27}) is acceptable because the inequality holds for all (x) in the open interval.But wait, if (a = frac{1}{27}), then at (x = frac{1}{3}), (f(x) = 0), but since (x = frac{1}{3}) is not included, it's fine. So, (a = frac{1}{27}) is acceptable. Therefore, the interval should include (frac{1}{27}), making it ([frac{1}{27}, 1)). But since 1 is excluded, the correct interval is ([frac{1}{27}, 1)), but that's not an option. The closest option is A: ([frac{1}{27}, 1]), but since (a) cannot be 1, maybe the answer is A, considering that 1 is not actually attainable, but the options don't have a half-open interval ending at 1.Wait, the options are:A: ([frac{1}{27}, 1])B: ((frac{1}{27}, 1))C: ((0, frac{1}{27}))D: ((0, frac{1}{27}])So, since (a) must be greater than (frac{1}{27}) and less than 1, the correct interval is ((frac{1}{27}, 1)), which is option B.But earlier, I thought that (a = frac{1}{27}) is acceptable because the inequality holds for all (x) in the open interval. So, why is the answer not including (frac{1}{27})?Wait, perhaps I made a mistake in assuming that (a = frac{1}{27}) is acceptable. Let me re-examine that.If (a = frac{1}{27}), then at (x = frac{1}{3}), (f(x) = 0), but since (x = frac{1}{3}) is not in the interval, it's okay. However, for (a = frac{1}{27}), is (f(x) < 0) for all (x in (0, frac{1}{3}))?Yes, because as (x) approaches (frac{1}{3}), (f(x)) approaches 0 from below, and for all (x) less than (frac{1}{3}), (f(x)) is less than 0. Therefore, (a = frac{1}{27}) is acceptable.But then, why is the answer not including 1? Because (a) cannot be 1, as (log_1 x) is undefined. So, the interval should be ([frac{1}{27}, 1)), but since that's not an option, the closest is A: ([frac{1}{27}, 1]), but technically, 1 is excluded. However, in the options, A includes 1, which is not allowed, while B is ((frac{1}{27}, 1)), which excludes both endpoints. But since (a = frac{1}{27}) is acceptable, the correct interval should include (frac{1}{27}), making the answer A.Wait, but the problem says "the inequality holds true for any (x in (0, frac{1}{3}))". So, if (a = frac{1}{27}), does the inequality hold for all (x) in that interval? Yes, because (f(x) < 0) for all (x) in ((0, frac{1}{3})). Therefore, (a = frac{1}{27}) is acceptable.But then, why is the answer not including 1? Because (a) cannot be 1. So, the correct interval is ([frac{1}{27}, 1)), but since that's not an option, perhaps the answer is A: ([frac{1}{27}, 1]), assuming that 1 is not actually attainable, but the options don't reflect that.Alternatively, maybe I made a mistake in considering (a = frac{1}{27}). Let me test (a = frac{1}{27}) with a specific (x) value.Let me choose (x = frac{1}{3}), but since (x) is not in the interval, I'll choose (x = frac{1}{4}).At (x = frac{1}{4}), (f(x) = 3 times (frac{1}{4})^2 - log_{1/27} (frac{1}{4}))Calculate:(3 times frac{1}{16} = frac{3}{16})(log_{1/27} (frac{1}{4}) = frac{ln (frac{1}{4})}{ln (frac{1}{27})} = frac{-ln 4}{-3 ln 3} = frac{ln 4}{3 ln 3} approx frac{1.386}{3 times 1.0986} approx frac{1.386}{3.2958} approx 0.420)So,(f(x) = frac{3}{16} - 0.420 approx 0.1875 - 0.420 = -0.2325 < 0)So, it's negative. Similarly, at (x = frac{1}{10}):(f(x) = 3 times (frac{1}{10})^2 - log_{1/27} (frac{1}{10}) = 0.03 - frac{ln (frac{1}{10})}{ln (frac{1}{27})} = 0.03 - frac{-ln 10}{-3 ln 3} = 0.03 - frac{ln 10}{3 ln 3})Calculate:(ln 10 approx 2.3026), (ln 3 approx 1.0986)So,(frac{2.3026}{3 times 1.0986} approx frac{2.3026}{3.2958} approx 0.700)Thus,(f(x) = 0.03 - 0.700 = -0.670 < 0)So, it's negative. Therefore, (a = frac{1}{27}) is acceptable because (f(x) < 0) for all (x in (0, frac{1}{3})).Therefore, the range of (a) is (frac{1}{27} leq a < 1). Since the options don't include a half-open interval ending at 1, but option A includes 1, which is not allowed, I think the correct answer is A: ([frac{1}{27}, 1]), assuming that 1 is not actually attainable, but the options don't reflect that. Alternatively, if the problem allows (a) to approach 1 but not reach it, then the interval is ([frac{1}{27}, 1)), but since that's not an option, the closest is A.Wait, but in the initial analysis, when (a = frac{1}{27}), the inequality holds for all (x) in the open interval, so it's acceptable. Therefore, the correct interval is ([frac{1}{27}, 1)), but since 1 is excluded, and the options don't have that, perhaps the answer is A, considering that 1 is not actually attainable, but the options don't reflect that.Alternatively, maybe the answer is A because the problem allows (a) to be up to 1, but in reality, (a) cannot be 1. However, since the options don't have a half-open interval, I think the intended answer is A: ([frac{1}{27}, 1]), even though technically 1 is excluded.But wait, looking back, when (a = 1), the logarithm is undefined, so (a) cannot be 1. Therefore, the correct interval should be ([frac{1}{27}, 1)), but since that's not an option, the closest is A: ([frac{1}{27}, 1]), but I think the intended answer is A.However, considering the options, and since (a = frac{1}{27}) is acceptable, and (a) must be less than 1, the correct interval is ([frac{1}{27}, 1)), but since it's not an option, perhaps the answer is A.Wait, no, the options are:A: ([frac{1}{27}, 1])B: ((frac{1}{27}, 1))C: ((0, frac{1}{27}))D: ((0, frac{1}{27}])So, since (a) must be greater than (frac{1}{27}) and less than 1, the correct interval is ((frac{1}{27}, 1)), which is option B.But earlier, I thought that (a = frac{1}{27}) is acceptable because the inequality holds for all (x) in the open interval. So, why is the answer not including (frac{1}{27})?Wait, perhaps I made a mistake in the initial analysis. Let me re-examine the inequality at (a = frac{1}{27}).At (a = frac{1}{27}), the inequality becomes (3x^2 - log_{1/27} x < 0).We can rewrite (log_{1/27} x = frac{ln x}{ln (1/27)} = frac{ln x}{-3 ln 3} = -frac{ln x}{3 ln 3}).So, the inequality becomes:(3x^2 - left(-frac{ln x}{3 ln 3}right) < 0)Simplify:(3x^2 + frac{ln x}{3 ln 3} < 0)But (ln x) is negative for (x in (0, 1)), so (frac{ln x}{3 ln 3}) is negative. Therefore, the inequality is:(3x^2 + text{(negative)} < 0)So, (3x^2) is positive, and we're adding a negative number. Whether the sum is negative depends on the magnitude.Wait, this seems different from before. Maybe I made a mistake in the earlier steps.Let me re-express the inequality:(3x^2 < log_a x)At (a = frac{1}{27}), (log_{1/27} x = frac{ln x}{ln (1/27)} = frac{ln x}{-3 ln 3} = -frac{ln x}{3 ln 3})So, the inequality becomes:(3x^2 < -frac{ln x}{3 ln 3})Multiply both sides by (3 ln 3) (which is positive because (ln 3 > 0)):(9x^2 ln 3 < -ln x)Multiply both sides by -1 (which reverses the inequality):(-9x^2 ln 3 > ln x)But (ln x) is negative, so let me write it as:(ln x < -9x^2 ln 3)So, we need (ln x < -9x^2 ln 3) for all (x in (0, frac{1}{3})).Let me check at (x = frac{1}{3}):(ln (frac{1}{3}) = -ln 3)(-9 (frac{1}{3})^2 ln 3 = -9 times frac{1}{9} ln 3 = -ln 3)So, equality holds at (x = frac{1}{3}). But since (x = frac{1}{3}) is not included, we need to check for (x) approaching (frac{1}{3}) from the left.As (x) approaches (frac{1}{3}), (ln x) approaches (-ln 3), and (-9x^2 ln 3) approaches (-9 times frac{1}{9} ln 3 = -ln 3). So, near (x = frac{1}{3}), (ln x) is slightly less than (-ln 3), and (-9x^2 ln 3) is slightly less than (-ln 3). Wait, no, because as (x) approaches (frac{1}{3}) from the left, (x^2) approaches (frac{1}{9}), so (-9x^2 ln 3) approaches (-ln 3). Therefore, (ln x) is slightly less than (-ln 3), so (ln x < -9x^2 ln 3) is true because (ln x) is more negative.Wait, no. Let me think again. If (x) is slightly less than (frac{1}{3}), say (x = frac{1}{3} - epsilon), then (ln x) is slightly less than (-ln 3), and (-9x^2 ln 3) is slightly less than (-ln 3). So, which one is more negative?Let me compute the difference:(ln x - (-9x^2 ln 3) = ln x + 9x^2 ln 3)At (x = frac{1}{3}), this is (-ln 3 + 9 times frac{1}{9} ln 3 = -ln 3 + ln 3 = 0).For (x) slightly less than (frac{1}{3}), let's approximate:Let (x = frac{1}{3} - epsilon), where (epsilon) is small.Then,(ln x approx ln (frac{1}{3}) - frac{epsilon}{frac{1}{3}} = -ln 3 - 3epsilon)And,(9x^2 ln 3 approx 9 (frac{1}{3})^2 ln 3 - 2 times 9 times frac{1}{3} times epsilon ln 3 = ln 3 - 6 epsilon ln 3)So,(ln x + 9x^2 ln 3 approx (-ln 3 - 3epsilon) + (ln 3 - 6 epsilon ln 3) = -3epsilon -6 epsilon ln 3)Since (epsilon > 0), this is negative. Therefore,(ln x + 9x^2 ln 3 < 0)Which implies,(ln x < -9x^2 ln 3)Therefore, the inequality holds for (x) near (frac{1}{3}). So, (a = frac{1}{27}) is acceptable.Therefore, the range of (a) is (frac{1}{27} leq a < 1). Since the options don't include a half-open interval ending at 1, but option A includes 1, which is not allowed, I think the correct answer is A: ([frac{1}{27}, 1]), assuming that 1 is not actually attainable, but the options don't reflect that.However, considering the options, and since (a = frac{1}{27}) is acceptable, and (a) must be less than 1, the correct interval is ([frac{1}{27}, 1)), but since that's not an option, the closest is A: ([frac{1}{27}, 1]).But wait, the problem says "the inequality holds true for any (x in (0, frac{1}{3}))". So, if (a = frac{1}{27}), the inequality holds for all (x) in that interval, so it's acceptable. Therefore, the range is ([frac{1}{27}, 1)), but since 1 is excluded, and the options don't have that, the answer must be A: ([frac{1}{27}, 1]), even though technically 1 is excluded.Alternatively, perhaps the answer is A because the problem allows (a) to be up to 1, but in reality, (a) cannot be 1. However, since the options don't have a half-open interval, I think the intended answer is A.But wait, looking back at the initial analysis, when (a = frac{1}{27}), the inequality holds for all (x) in the open interval, so it's acceptable. Therefore, the correct interval is ([frac{1}{27}, 1)), but since that's not an option, the closest is A: ([frac{1}{27}, 1]).However, considering that (a) cannot be 1, the correct answer should be B: ((frac{1}{27}, 1)), because (a = frac{1}{27}) is acceptable, but (a = 1) is not. Therefore, the interval is ((frac{1}{27}, 1)), which is option B.Wait, but earlier, I thought that (a = frac{1}{27}) is acceptable because the inequality holds for all (x) in the open interval. So, why is the answer not including (frac{1}{27})?I think I'm overcomplicating this. Let me summarize:- (a > 1): Inequality doesn't hold.- (0 < a < 1): Need (3x^2 < log_a x) for all (x in (0, frac{1}{3})).- At (x = frac{1}{3}), the inequality becomes (3 times frac{1}{9} < log_a frac{1}{3}), which simplifies to (frac{1}{3} < log_a frac{1}{3}).- Solving this gives (a > frac{1}{27}).- Since (a) must be less than 1, the interval is ((frac{1}{27}, 1)).- However, testing (a = frac{1}{27}) shows that the inequality holds for all (x) in the open interval, so (a = frac{1}{27}) is acceptable.- Therefore, the interval should be ([frac{1}{27}, 1)), but since that's not an option, the closest is A: ([frac{1}{27}, 1]).But since (a = 1) is not allowed, the correct answer should be B: ((frac{1}{27}, 1)).Wait, no. Because (a = frac{1}{27}) is acceptable, so the interval should include it. Therefore, the correct answer is A: ([frac{1}{27}, 1]).But I'm still confused because (a = 1) is not allowed. However, since the options don't have a half-open interval, I think the intended answer is A.But to be precise, since (a = 1) is not allowed, the correct interval is ([frac{1}{27}, 1)), which is not an option. Therefore, the closest is A: ([frac{1}{27}, 1]), but technically, it's not entirely correct.Alternatively, perhaps the answer is A because the problem allows (a) to approach 1 but not reach it, and since the options don't have a half-open interval, A is the best choice.But I think the correct answer is A: ([frac{1}{27}, 1]), considering that (a = frac{1}{27}) is acceptable and (a) approaches 1 but doesn't include it, even though the option includes 1.Wait, no. The problem doesn't specify whether (a) can be 1 or not, but since (log_1 x) is undefined, (a) cannot be 1. Therefore, the correct interval is ([frac{1}{27}, 1)), but since that's not an option, the answer must be A: ([frac{1}{27}, 1]), assuming that 1 is not actually attainable, but the options don't reflect that.I think I've spent too much time on this, but I'll go with A: ([frac{1}{27}, 1]), because (a = frac{1}{27}) is acceptable, and the problem might consider the interval up to 1, even though technically 1 is excluded.

💡First, I need to determine the original price of the article before any concessions were applied.Arun purchased the article for ₹1,120 after receiving two successive discounts: a 30% standard concession followed by an additional 20% concession on the reduced price.Let’s denote the original price of the article as P.After the first 30% discount, the price becomes 70% of the original price:P × 0.70Then, an additional 20% discount is applied to this reduced price, making the final price 80% of the already discounted price:0.70P × 0.80 = 0.56PSince Arun paid ₹1,120 for the article, I can set up the equation:0.56P = 1,120To find the original price P, I will divide both sides of the equation by 0.56:P = 1,120 / 0.56P = 2,000Therefore, the original price of the article was ₹2,000.

💡Okay, so I have this problem here where I need to find f'(1) given the function f(x) = 2x f'(1) + ln x. Hmm, that looks a bit tricky because f'(1) is inside the function itself. I guess I need to figure out a way to solve for f'(1). Let me think about how to approach this.First, I know that f'(x) is the derivative of f(x) with respect to x. So maybe I should start by finding the derivative of f(x). Let me write that down. The function is f(x) = 2x f'(1) + ln x. So, to find f'(x), I need to differentiate each term separately.The first term is 2x f'(1). Since f'(1) is a constant (because it's the derivative evaluated at x=1), the derivative of 2x f'(1) with respect to x is just 2 f'(1). That's straightforward.The second term is ln x. The derivative of ln x with respect to x is 1/x. So, putting it all together, f'(x) should be 2 f'(1) + 1/x. Okay, that makes sense.Now, the problem asks for f'(1). So I need to evaluate f'(x) at x=1. Let me substitute x=1 into the derivative I just found. That gives me f'(1) = 2 f'(1) + 1/1. Simplifying that, it becomes f'(1) = 2 f'(1) + 1.Hmm, now I have an equation where f'(1) is on both sides. I need to solve for f'(1). Let me rearrange the equation. If I subtract 2 f'(1) from both sides, I get f'(1) - 2 f'(1) = 1. That simplifies to -f'(1) = 1. So, multiplying both sides by -1, I get f'(1) = -1.Wait, does that make sense? Let me double-check my steps. I took the derivative of f(x), which was 2x f'(1) + ln x, and got f'(x) = 2 f'(1) + 1/x. Then I plugged in x=1, so f'(1) = 2 f'(1) + 1. Solving for f'(1), I subtracted 2 f'(1) from both sides, leading to -f'(1) = 1, so f'(1) = -1. Yeah, that seems correct.I guess the key here was recognizing that f'(1) is a constant, so when taking the derivative, it just becomes a coefficient. Then, by plugging in x=1 into the derivative, I could set up an equation to solve for f'(1). It's interesting how the function is defined in terms of its own derivative at a specific point. That makes it a bit of a self-referential problem, but it's solvable by differentiating and then substituting.I wonder if there's another way to approach this problem. Maybe by integrating or something else? Let me think. If I try to write the function f(x) as 2x f'(1) + ln x, and then integrate f'(x), but that might complicate things because I don't know f'(1) yet. It seems like differentiating was the more straightforward approach.Another thing I could check is whether the function f(x) satisfies the original equation when f'(1) is -1. Let me substitute f'(1) = -1 back into the original function and see if everything holds.So, f(x) = 2x*(-1) + ln x = -2x + ln x. Now, let's compute f'(x) for this function. The derivative of -2x is -2, and the derivative of ln x is 1/x. So f'(x) = -2 + 1/x. Now, evaluating f'(1), we get f'(1) = -2 + 1/1 = -2 + 1 = -1. Perfect, that matches our earlier result. So f'(1) = -1 is indeed correct.I think I'm confident with this answer now. It all checks out when I substitute back into the original function and verify the derivative. It's a neat problem because it shows how sometimes you can solve for unknown constants by using the properties of derivatives and substitution.Just to recap, the steps I took were:1. Start with the given function f(x) = 2x f'(1) + ln x.2. Differentiate f(x) with respect to x to find f'(x).3. Substitute x=1 into the derivative to get an equation involving f'(1).4. Solve the resulting equation for f'(1).5. Verify the solution by substituting back into the original function and checking the derivative.This systematic approach helped me avoid confusion and ensure that my solution was correct. It's a good reminder that even when a problem seems a bit circular or self-referential, breaking it down step by step can lead to a clear solution.I also thought about whether there might be alternative methods, like integrating or using differential equations, but in this case, differentiation was the most direct path. It's a good example of how knowing the fundamental rules of calculus can help solve problems that might initially seem tricky.Overall, I feel more comfortable now with problems where the function is defined in terms of its own derivative at a specific point. It's all about recognizing that the derivative at that point is a constant and then using substitution to solve for it.

💡Alright, so I have this problem where I need to find the coefficient of the ( x^2 ) term in the expansion of the binomial ( left(a sqrt{x} - frac{1}{sqrt{x}}right)^6 ). The value of ( a ) comes from some flowchart, and it's given as ( a ). Hmm, okay, I guess I need to figure out what ( a ) is first. Wait, the problem mentions a flowchart, but I don't have the diagram here. Maybe ( a ) is determined by some cyclic process? The user mentioned something about a cycle with values 2, -1, and ( frac{1}{2} ). Oh, right, for ( i = 2016 ), the value of ( a ) is ( frac{1}{2} ). Okay, so I can take ( a = frac{1}{2} ) for this problem.So, substituting ( a = frac{1}{2} ) into the binomial expression, it becomes ( left(frac{1}{2} sqrt{x} - frac{1}{sqrt{x}}right)^6 ). Now, I need to expand this and find the coefficient of ( x^2 ).I remember the binomial theorem, which says that ( (x + y)^n = sum_{k=0}^{n} binom{n}{k} x^{n - k} y^k ). So, applying this to our expression, each term in the expansion will be of the form ( binom{6}{k} left(frac{1}{2} sqrt{x}right)^{6 - k} left(-frac{1}{sqrt{x}}right)^k ).Let me write that out more clearly:[left(frac{1}{2} sqrt{x} - frac{1}{sqrt{x}}right)^6 = sum_{k=0}^{6} binom{6}{k} left(frac{1}{2} sqrt{x}right)^{6 - k} left(-frac{1}{sqrt{x}}right)^k]Now, I need to simplify each term. Let's break it down:1. The binomial coefficient is ( binom{6}{k} ).2. The first part is ( left(frac{1}{2} sqrt{x}right)^{6 - k} ). This can be written as ( left(frac{1}{2}right)^{6 - k} times (sqrt{x})^{6 - k} ).3. The second part is ( left(-frac{1}{sqrt{x}}right)^k ). This can be written as ( (-1)^k times left(frac{1}{sqrt{x}}right)^k ).Putting it all together, each term becomes:[binom{6}{k} times left(frac{1}{2}right)^{6 - k} times (sqrt{x})^{6 - k} times (-1)^k times left(frac{1}{sqrt{x}}right)^k]Simplify the exponents of ( x ):- ( (sqrt{x})^{6 - k} = x^{frac{6 - k}{2}} )- ( left(frac{1}{sqrt{x}}right)^k = x^{-frac{k}{2}} )Multiplying these together:[x^{frac{6 - k}{2}} times x^{-frac{k}{2}} = x^{frac{6 - k - k}{2}} = x^{frac{6 - 2k}{2}} = x^{3 - k}]So, each term simplifies to:[binom{6}{k} times left(frac{1}{2}right)^{6 - k} times (-1)^k times x^{3 - k}]We need the term where the exponent of ( x ) is 2. So, set ( 3 - k = 2 ):[3 - k = 2 implies k = 1]So, the term corresponding to ( k = 1 ) will have ( x^2 ). Let's compute that term:[binom{6}{1} times left(frac{1}{2}right)^{6 - 1} times (-1)^1 times x^{2}]Calculate each part:- ( binom{6}{1} = 6 )- ( left(frac{1}{2}right)^{5} = frac{1}{32} )- ( (-1)^1 = -1 )Putting it all together:[6 times frac{1}{32} times (-1) times x^2 = -frac{6}{32} x^2 = -frac{3}{16} x^2]So, the coefficient of ( x^2 ) is ( -frac{3}{16} ).Wait, let me double-check my calculations to make sure I didn't make a mistake. First, binomial coefficient ( binom{6}{1} ) is indeed 6. Then, ( left(frac{1}{2}right)^5 ) is ( frac{1}{32} ). Multiplying 6 and ( frac{1}{32} ) gives ( frac{6}{32} ), which simplifies to ( frac{3}{16} ). Then, multiplying by ( -1 ) gives ( -frac{3}{16} ). Yes, that seems correct. So, I think the coefficient is ( -frac{3}{16} ).**Final Answer**The coefficient of the ( x^2 ) term is boxed{-dfrac{3}{16}}.

💡To determine the imaginary part of the complex number ( z = frac{3 + i}{1 + i} ), we can follow these steps:1. **Multiply the numerator and denominator by the conjugate of the denominator** to eliminate the imaginary unit from the denominator: [ z = frac{(3 + i)(1 - i)}{(1 + i)(1 - i)} ]2. **Expand both the numerator and the denominator**: - Numerator: [ (3 + i)(1 - i) = 3 cdot 1 + 3 cdot (-i) + i cdot 1 + i cdot (-i) = 3 - 3i + i - i^2 ] Since ( i^2 = -1 ), this simplifies to: [ 3 - 3i + i - (-1) = 3 - 2i + 1 = 4 - 2i ] - Denominator: [ (1 + i)(1 - i) = 1 cdot 1 + 1 cdot (-i) + i cdot 1 + i cdot (-i) = 1 - i + i - i^2 ] Again, since ( i^2 = -1 ), this simplifies to: [ 1 - i + i - (-1) = 1 + 1 = 2 ]3. **Simplify the expression**: [ z = frac{4 - 2i}{2} = 2 - i ]4. **Identify the imaginary part** of the simplified complex number ( 2 - i ), which is ( -1 ).Therefore, the imaginary part of ( z ) is (-1).

💡Okay, so I have this algebra problem here, and I need to simplify this expression:[frac{P+Q}{P-Q} - frac{P-Q}{P+Q}]where ( P = 2x + 3y ) and ( Q = x - 2y ). Then, I have to choose the correct simplified form from the given options A to E.Alright, let's start by understanding what the problem is asking. It's an expression with two fractions being subtracted. Both fractions have denominators that involve ( P ) and ( Q ), and numerators that are also combinations of ( P ) and ( Q ). So, my first thought is that I need to substitute the given expressions for ( P ) and ( Q ) into this main expression and then simplify it step by step.First, let me write down what ( P + Q ) and ( P - Q ) are. Since ( P = 2x + 3y ) and ( Q = x - 2y ), adding them together:( P + Q = (2x + 3y) + (x - 2y) )Let me compute that:( 2x + x = 3x )( 3y - 2y = y )So, ( P + Q = 3x + y )Similarly, subtracting ( Q ) from ( P ):( P - Q = (2x + 3y) - (x - 2y) )Let me compute that:( 2x - x = x )( 3y - (-2y) = 3y + 2y = 5y )So, ( P - Q = x + 5y )Alright, so now I can rewrite the original expression with these simplified terms:[frac{3x + y}{x + 5y} - frac{x + 5y}{3x + y}]Now, I need to subtract these two fractions. To subtract fractions, they need a common denominator. The denominators here are ( x + 5y ) and ( 3x + y ). The common denominator would be the product of these two, which is ( (x + 5y)(3x + y) ).So, I can rewrite each fraction with this common denominator:First fraction: ( frac{3x + y}{x + 5y} ) becomes ( frac{(3x + y)(3x + y)}{(x + 5y)(3x + y)} )Second fraction: ( frac{x + 5y}{3x + y} ) becomes ( frac{(x + 5y)(x + 5y)}{(x + 5y)(3x + y)} )Wait, actually, that's not quite right. When you have a fraction ( frac{a}{b} ) and you want to express it with denominator ( bc ), you multiply numerator and denominator by ( c ). So, for the first fraction, it's:( frac{3x + y}{x + 5y} = frac{(3x + y)(3x + y)}{(x + 5y)(3x + y)} )Similarly, the second fraction:( frac{x + 5y}{3x + y} = frac{(x + 5y)(x + 5y)}{(x + 5y)(3x + y)} )So, now both fractions have the same denominator, and I can subtract the numerators:[frac{(3x + y)^2 - (x + 5y)^2}{(x + 5y)(3x + y)}]Alright, so now I need to compute ( (3x + y)^2 ) and ( (x + 5y)^2 ), subtract them, and then simplify the numerator.Let me compute ( (3x + y)^2 ):( (3x + y)^2 = (3x)^2 + 2*(3x)*(y) + y^2 = 9x^2 + 6xy + y^2 )Similarly, ( (x + 5y)^2 ):( (x + 5y)^2 = x^2 + 2*x*(5y) + (5y)^2 = x^2 + 10xy + 25y^2 )Now, subtracting these two:( (9x^2 + 6xy + y^2) - (x^2 + 10xy + 25y^2) )Let me distribute the negative sign:( 9x^2 + 6xy + y^2 - x^2 - 10xy - 25y^2 )Now, combine like terms:- ( 9x^2 - x^2 = 8x^2 )- ( 6xy - 10xy = -4xy )- ( y^2 - 25y^2 = -24y^2 )So, the numerator simplifies to:( 8x^2 - 4xy - 24y^2 )Alright, now the denominator is ( (x + 5y)(3x + y) ). Let me compute that as well to see if it can be simplified or factored further.Multiplying ( (x + 5y)(3x + y) ):First, multiply ( x * 3x = 3x^2 )Then, ( x * y = xy )Then, ( 5y * 3x = 15xy )Finally, ( 5y * y = 5y^2 )Adding these up:( 3x^2 + xy + 15xy + 5y^2 )Combine like terms:( 3x^2 + (xy + 15xy) + 5y^2 = 3x^2 + 16xy + 5y^2 )So, the denominator is ( 3x^2 + 16xy + 5y^2 )Putting it all together, the entire expression simplifies to:[frac{8x^2 - 4xy - 24y^2}{3x^2 + 16xy + 5y^2}]Now, let me look at the answer choices:A) ( frac{8x^2 - 4xy - 24y^2}{3x^2 + 16xy + 5y^2} )B) ( frac{8x^2 + 4xy - 24y^2}{3x^2 + 16xy + 5y^2} )C) ( frac{8x^2 - 4xy + 24y^2}{3x^2 + 16xy + 5y^2} )D) ( frac{8x^2 + 24y^2}{3x^2 + 16xy + 5y^2} )E) ( frac{-8x^2 + 24y^2}{3x^2 + 16xy + 5y^2} )Comparing my result with the options, it matches option A exactly.Wait, just to make sure I didn't make any mistakes in my calculations, let me double-check each step.First, substituting ( P ) and ( Q ):( P + Q = 2x + 3y + x - 2y = 3x + y ) – correct.( P - Q = 2x + 3y - (x - 2y) = 2x + 3y - x + 2y = x + 5y ) – correct.Then, rewriting the expression:[frac{3x + y}{x + 5y} - frac{x + 5y}{3x + y}]Finding a common denominator: ( (x + 5y)(3x + y) ) – correct.Expanding the numerators:( (3x + y)^2 = 9x^2 + 6xy + y^2 ) – correct.( (x + 5y)^2 = x^2 + 10xy + 25y^2 ) – correct.Subtracting them:( 9x^2 + 6xy + y^2 - x^2 - 10xy - 25y^2 = 8x^2 - 4xy - 24y^2 ) – correct.Denominator:( (x + 5y)(3x + y) = 3x^2 + xy + 15xy + 5y^2 = 3x^2 + 16xy + 5y^2 ) – correct.So, the simplified expression is indeed:[frac{8x^2 - 4xy - 24y^2}{3x^2 + 16xy + 5y^2}]Which is option A.I think I did everything correctly. I don't see any mistakes in my steps. So, I'm confident that option A is the correct answer.**Final Answer**boxed{A}